This is called Modular Exponentiation and can be solved in many different ways. The most efficient is Right-to-Left Binary Method (on Wikipedia). It is possible to do by hand and will only take 8 iterations of the algorithm, but may be tedious.

Also by removing the gcd of 7 from the numerator and denominator you are essentially changing the question. The remainder is now being limited and will not be the same.
Essentially: $$7^{369}\mod{350} \neq 7^{368}\mod{50}$$
and $$7^{369}\mod{350} \neq 1$$
however if you multipy by the gcd that you divided by in the beginning then you will get the correct answer
$$7^{369}\mod{350}=7*(7^{368}\mod{50}) = 7$$

$\begingroup$I think this answer is in the right track$\endgroup$
– HaileyJun 7 '16 at 8:56

$\begingroup$@Hailey Thanks, everyone else hasn't realized that by reducing the denominator you are also changing the remainder. Like 16/6 = 8/3 but the remainder for 16/6 = 4 and the remainder for 8/3 = 2.$\endgroup$
– Noah GerenJun 7 '16 at 9:01

$\begingroup$thats why after getting 1,we have to multiply with 7 to get 7$\endgroup$
– HaileyJun 7 '16 at 10:47

One thing you can sometimes do with problems like this is to find the remainder modulo the different prime powers and then piece the information together to find the original remainder. Essentially this amounts to using the Chinese Remainder Theorem which you may not have learned yet but one doesn't really need to know it here.

Since $350 = 2\cdot 5^2 \cdot 7$ we compute
$$ 7^{369} \equiv 1^{369}=1 \pmod{2} $$
and
$$ 7^{369} = 7\cdot (7^2)^{184} \equiv 7\cdot(-1)^{184} = 7\cdot 1 = 7\pmod{5^2}$$
and
$$ 7^{369} \equiv 0^{369}=0\pmod{7}.$$
Now only possible remainders $r< 350$ sastisfying $r\equiv 7 \pmod{5^2}$ are $7, 25+7, \ldots, 325+7$. Since we must also have $r\equiv 0 \pmod{7}$ we are only left with $r=7$ or $r=7\cdot 25+7=182$. Finally using $r\equiv 1\pmod{2}$, we see that $r=7$.